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Krakow, Summer 2011 Circle and Sphere Orders William T. Trotter

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Presentation on theme: "Krakow, Summer 2011 Circle and Sphere Orders William T. Trotter"— Presentation transcript:

1 Krakow, Summer 2011 Circle and Sphere Orders William T. Trotter trotter@math.gatech.edu

2 Families of Disks

3 Contact Graphs

4 A Theorem with Many Proofs Theorem (Koebe 1936, Andreev 1970, Thurston 1985, and many others) Every planar graph is the contact graph of a family of circular disks in the plane. Note All known proofs are non-constructive.

5 Circle Orders

6 Circle and Sphere Orders Question Which posets are circle orders? Question For each d, which posets can be represented as inclusion orders of d- dimensional spheres? Conjecture Every 3-dimensional poset is a circle order.

7 Degrees of Freedom – Informal Discussion Definition A family F of posets has (at most) k degrees of freedom when it is possible to assign to each element x in a poset in F a sequence x = (x 1, x 2, …, x k ) of real numbers so that there is a finite family p 1, p 2, …,p r of polynomials in 2k variables so that the issue of whether x ≤ y in P is determined by the sign pattern of p 1 (x,y), p 2 (x,y),…, p r (x,y).

8 Degrees of Freedom – Examples Examples Interval orders and interval inclusion orders both have two degrees of freedom. Exercise Circle orders have three degrees of freedom. More generally, spheres in R d have d + 1 degrees of freedom.

9 Alon and Scheinerman’s Theorem Theorem (Alon and Scheinerman) For every positive integer k, when n is large relative to k, almost no posets on n points having dimension at least k + 1 have a representation by a family which has only k degrees of freedom. Remark This theorem is proven by first showing an lower bound on the number of posets on n points having dimension at most k + 1. Then a theorem of Warren is used to establish an upper bound for posets representable with only k degrees of freedom.

10 Sometimes a Matter of Overkill Remark By the degrees of freedom argument, when n is large, almost all posets of dimension 4 on n points are not circle orders. Exercise Show that the 14 element poset obtained by removing the zero and the one from the 4-dimensional subset lattice 2 4 is not a circle order. This is the reason why Venn diagrams are not a useful tool in discussions of intersections and unions of four sets.

11 Some Basic Results Remark Spheres in 1-space are just intervals of R, the set of real numbers. A poset P is the inclusion order of a family of intervals of R if and only if dim(P) ≤ 2. Observation Since circle orders have 3 degrees of freedom, this makes it seem reasonable that all 3-dimensional posets are circle orders.

12 Motivating Results Theorem (Schnyder) A graph G is planar if and only if the dimension of its incidence poset (vertices and edges ordered by inclusion) has dimension at most 3. Theorem (Brightwell and Scheinerman) A graph G is planar if and only if its incidence poset is a circle order.

13 More Results for Circle Orders Exercise Show that for each n, the standard example S n is a circle order. Theorem (Fishburn) Every interval order is a circle order.

14 The Answer Should be “Yes” Exercise Let P be a poset with dim(P) ≤ 3. P should be a circle order because for every n ≥ 3, P can be represented as the inclusion order of regular n - gons in the plane (of variable size but with a fixed orientation). When n is very large, a regular n - gon is essentially just a circle.

15 The Answer Should be “No” Theorem (Scheinerman and Weirman) The countably infinite 3-dimensional poset Z × Z × Z is not a circle order. Theorem (Fon der Flaass) The countably infinite poset 2 × 3 × N is not a sphere order. Remark These results are for infinite posets. But perhaps it is still true that every finite 3- dimensional poset is a circle order.

16 The Real Answer Theorem (Felsner, Fishburn and Trotter) When n is sufficiently large, the finite 3- dimensional poset n × n × n is not a sphere order. Remark The integer n in the proof of the preceding theorem is huge, but in fact, the result might hold when n is very reasonable, perhaps even when n < 10.


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